The problem can be addressed using macroscopic thermodynamics without referring to the molecular picture of the fluid. We however need to know something about the chemical composition of the fluid after the reaction. I suppose that not all of your fluid undergoes the chemical reaction, so the resulting fluid is going to be multi-component. Below, I sketch a solution of the problem under the assumption that all the components move with the same macroscopic velocity, so one can treat the fluid after the reaction as a single-component matter with some average thermodynamic properties. A more detailed solution would depend on the exact nature of your fluid, e.g. on whether the ideal-gas approximation is appropriate.
The state of the fluid is completely described by its temperature $T$, pressure $P$, density $\rho$ and velocity $v$. I assume, as you indicated, that the geometry of the problem is one-dimensional so that the velocity can be treated as a scalar. I will use indices 1 and 2 to refer to the fluid before and after the reaction. There is one relation between $T$, $P$ and $\rho$ supplied by the equation of state, so we need three additional relations to fully determine the unknown values of $T_2$, $P_2$, $\rho_2$ and $v_2$.
Mass conservation. Given that the cross-section area of the flow does not change, this amounts to the simple condition
$$
\rho_1v_1=\rho_2v_2,
$$
equivalent to your equation.
Momentum balance. The flow of momentum per unit area and per second in a fluid of velocity $v$ is $\rho v^2$, which is just the mass flow times velocity. Then the second law of Newton requires that
$$
P_1-P_2=\rho_2v_2^2-\rho_1v_1^2.
$$
Energy balance. The above two constraints are purely mechanical. Here is the only place where we need some input about the chemical reaction in the system. I will model the reaction as an instantaneous supply of heat per kilogram equal to $q$. Then the energy balance condition can be expressed in terms of increase of specific enthalpy $h$ (enthalpy per kilogram) plus specific kinetic energy (which is just $\frac12v^2$),
$$
h_2+\frac12v_2^2=h_1+\frac12v_1^2+q.
$$
The enthalpy itself is in turn determined by the temperature and pressure. For an ideal gas, it can be extracted for instance from the knowledge of specific heat at constant pressure.
The above set of equations together with the equation of state is sufficient to determine the final state of your fluid. As said, a more concrete solution can only be given in case you know the equation of state of your fluid. Hope this helps anyway.
EDIT. In classical thermodynamics, thermodynamic potentials such as enthalpy are only defined up to an arbitrary additive constant, which can be chosen independently for different materials. In order that the comparison of $h_1$ and $h_2$ makes sense, we therefore have to fix the reference values of the enthalpy of the fluid before and after the reaction appropriately. In my answer, I assume that the specific enthalpies of these two different fluids are equal at temperature $T_1$ and pressure $P_1$.
Best Answer
$\rho_1A_1V_1=\rho_2A_2V_2$ is used when the fluid is compressible, i.e., density at different points of fluid can also be different. In contrast, $A_1V_1=A_2V_2$ is used when the fluid is incompressible, i.e., its density is constant for every point in the fluid.